实验5

实验任务1

#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
void find_min_max(int x[], int n, int *pmin, int *pmax);

int main() {
    int a[N];
    int min, max;

    printf("录入%d个数据:\n", N);
    input(a, N);

    printf("数据是: \n");
    output(a, N);

    printf("数据处理...\n");
    find_min_max(a, N, &min, &max);

    printf("输出结果:\n");
    printf("min = %d, max = %d\n", min, max);

    return 0;
}

void input(int x[], int n) {
    int i;

    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}

void output(int x[], int n) {
    int i;
    
    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

void find_min_max(int x[], int n, int *pmin, int *pmax) {
    int i;
    
    *pmin = *pmax = x[0];

    for(i = 0; i < n; ++i)
        if(x[i] < *pmin)
            *pmin = x[i];
        else if(x[i] > *pmax)
            *pmax = x[i];
}

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1)找出一组数据中的最大和最小值;

2)指向x[0]

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
int *find_max(int x[], int n);

int main() {
    int a[N];
    int *pmax;

    printf("录入%d个数据:\n", N);
    input(a, N);

    printf("数据是: \n");
    output(a, N);

    printf("数据处理...\n");
    pmax = find_max(a, N);

    printf("输出结果:\n");
    printf("max = %d\n", *pmax);

    return 0;
}

void input(int x[], int n) {
    int i;

    for(i = 0; i < n; ++i)
        scanf("%d", &x[i]);
}

void output(int x[], int n) {
    int i;
    
    for(i = 0; i < n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

int *find_max(int x[], int n) {
    int max_index = 0;
    int i;

    for(i = 0; i < n; ++i)
        if(x[i] > x[max_index])
            max_index = i;
    
    return &x[max_index];
}

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 1)找出一组数据的最大值,并返回它的地址;

2)可以

实验任务2

#include <stdio.h>
#include <string.h>
#define N 80

int main() {
    char s1[N] = "Learning makes me happy";
    char s2[N] = "Learning makes me sleepy";
    char tmp[N];

    printf("sizeof(s1) vs. strlen(s1): \n");
    printf("sizeof(s1) = %d\n", sizeof(s1));
    printf("strlen(s1) = %d\n", strlen(s1));

    printf("\nbefore swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);

    printf("\nswapping...\n");
    strcpy(tmp, s1);
    strcpy(s1, s2);
    strcpy(s2, tmp);

    printf("\nafter swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);

    return 0;
}

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1)大小为80;计算数组所占字节数;统计字符串长度;

2)不能;s1是常量,不能被赋值;

3)交换了

#include <stdio.h>
#include <string.h>
#define N 80

int main() {
    char *s1 = "Learning makes me happy";
    char *s2 = "Learning makes me sleepy";
    char *tmp;

    printf("sizeof(s1) vs. strlen(s1): \n");
    printf("sizeof(s1) = %d\n", sizeof(s1));
    printf("strlen(s1) = %d\n", strlen(s1));

    printf("\nbefore swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);

    printf("\nswapping...\n");
    tmp = s1;
    s1 = s2;
    s2 = tmp;

    printf("\nafter swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);

    return 0;
}

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1)存放L的地址;计算指针所占字节数;统计指针所指字符串长度;

2)可以,1里是常量,2里是变量,后者可以被赋值;

3)交换了s1,s2的地址;内存中并未交换

实验任务3

#include <stdio.h>

int main() {
    int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
    int i, j;
    int *ptr1;     // 指针变量,存放int类型数据的地址
    int(*ptr2)[4]; // 指针变量,指向包含4个int元素的一维数组

    printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
    for (i = 0; i < 2; ++i) {
        for (j = 0; j < 4; ++j)
            printf("%d ", x[i][j]);
        printf("\n");
    }

    printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n");
    for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
        printf("%d ", *ptr1);

        if ((i + 1) % 4 == 0)
            printf("\n");
    }
                         
    printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n");
    for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
        for (j = 0; j < 4; ++j)
            printf("%d ", *(*ptr2 + j));
        printf("\n");
    }

    return 0;
}

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1)int (*ptr)[4]中,ptr指向一个包含4个int元素的数组;int *ptr[4]中,ptr表示一个包含四个int指针的数组;

实验任务4

#include <stdio.h>
#define N 80

void replace(char *str, char old_char, char new_char); // 函数声明

int main() {
    char text[N] = "Programming is difficult or not, it is a question.";

    printf("原始文本: \n");
    printf("%s\n", text);

    replace(text, 'i', '*'); // 函数调用 注意字符形参写法,单引号不能少

    printf("处理后文本: \n");
    printf("%s\n", text);

    return 0;
}

// 函数定义
void replace(char *str, char old_char, char new_char) {
    int i;

    while(*str) {
        if(*str == old_char)
            *str = new_char;
        str++;
    }
}

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1)将字符串中old_char替换为new_char;

2)可以

实验任务5

#include <stdio.h>
#define N 80

char* str_trunc(char* str, char x);

int main() {
    char str[N];
    char ch;

    while (printf("输入字符串: "), gets(str) != NULL) {
        printf("输入一个字符: ");
        ch = getchar();

        printf("截断处理...\n");
        str_trunc(str, ch);         // 函数调用

        printf("截断处理后的字符串: %s\n\n", str);
        getchar();
    }

    return 0;
}

char* str_trunc(char* str, char x)
{
    while (*str) {
         if (*str == x)
             * str = '\0';
         str++;
    }
}

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1)消除缓冲区的换行符

实验任务6

#include <stdio.h>
#include <string.h>
#define N 5

int check_id(char* str); // 函数声明

int main()
{
    char* pid[N] = { "31010120000721656X",
                    "3301061996X0203301",
                    "53010220051126571",
                    "510104199211197977",
                    "53010220051126133Y" };
    int i;

    for (i = 0; i < N; ++i)
        if (check_id(pid[i])) // 函数调用
            printf("%s\tTrue\n", pid[i]);
        else
            printf("%s\tFalse\n", pid[i]);

    return 0;
}

// 函数定义
// 功能: 检查指针str指向的身份证号码串形式上是否合法
// 形式合法,返回1,否则,返回0

int check_id(char* str)
{
    if (strlen(str) != 18)
        return 0;

    for (int i = 0; i < 17; ++i)
    {
        if (str[i] < '0' || str[i]>'9')
        {
            return 0;
        }
    }

    char id_18 = str[17];

    if (id_18 != 'X' && (str[17] < '0' || str[17]>'9'))
    {
        return 0;
    }
    else
        return 1;
}

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 实验任务7

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#define N 80
void encoder(char* str, int n); // 函数声明
void decoder(char* str, int n); // 函数声明

int main() {
    char words[N];
    int n;

    printf("输入英文文本: ");
    gets(words);

    printf("输入n: ");
    scanf("%d", &n);

    printf("编码后的英文文本: ");
    encoder(words, n);      // 函数调用
    printf("%s\n", words);

    printf("对编码后的英文文本解码: ");
    decoder(words, n); // 函数调用
    printf("%s\n", words);

    return 0;
}

/*函数定义
功能:对str指向的字符串进行编码处理
编码规则:
对于a~z或A~Z之间的字母字符,用其后第n个字符替换; 其它非字母字符,保持不变
*/
void encoder(char* str, int n) {
    while (*str) {
        if (*str >= 'a' && *str <= 'z') {
            *str = 'a' + (*str - 'a' + n) % 26;
        }
        else if (*str >= 'A' && *str <= 'Z') {
            *str = 'A' + (*str - 'A' + n) % 26;
        }
        str++;
    }
}

/*函数定义
功能:对str指向的字符串进行解码处理
解码规则:
对于a~z或A~Z之间的字母字符,用其前面第n个字符替换; 其它非字母字符,保持不变
*/
void decoder(char* str, int n) {
    while (*str) {
        if (*str >= 'a' && *str <= 'z') {
            *str = 'a' + (*str - 'a' - n + 26) % 26;
        }
        else if (*str >= 'A' && *str <= 'Z') {
            *str = 'A' + (*str - 'A' - n + 26) % 26;
        }
        str++;
    }
}

 

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 实验任务8

#include <stdio.h>
#include<string.h>

int main(int argc, char *argv[]) {
    int i,j,k;
    char *tmp;
    for(i=1;i<argc;++i)
    {
        for(j=1;j<argc-1;++j)
        {
            if(strcmp(*(argv+j),*(argv+j+1))>0)
            {
                tmp=*(argv+j);
                *(argv+j)=*(argv+j+1);
                *(argv+j+1)=tmp;
            }
        }
    }

    for(i = 1; i < argc; ++i)
        printf("hello, %s\n", argv[i]);

    return 0;
}

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posted @ 2025-12-12 15:49  Fandhi  阅读(0)  评论(0)    收藏  举报